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We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21-sided polygon.

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The sum of the interior angles of a polygon having 'n' sides is given by \left( n-2 \right)\ast 180^{\circ}

By this formula, the sum of angles with 3 sides, 4 sides, 5 sides and 6 sides is 180^{\circ},~360^{\circ},~540^{\circ},720^{\circ}

respectively. As the number of side increases by 1, the sum of interior angles increases by 180^{\circ}

So,for sum of angles of polygon with 21 sides = \left( n-2 \right) \ast180^{\circ}~ \\\\

\\ = \left( 21-2 \right) \ast 180^{\circ}=19\ast 180^{\circ}~ \\\\ =3420^{\circ} \\\\

 

 

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